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I Tangential Velocity w/r to change in Centripetal force

  1. Sep 12, 2016 #1
    Hi I have a scenario with which I wish to understand .

    I have a ball on a string which is in uniform circular rotation at Tangential velocity v
    Assume friction is not present
    upload_2016-9-13_11-25-16.png

    I now increase the centripetal force by pulling on the string from the centre,
    what has happened to the tangential velocity ?


    upload_2016-9-13_11-24-27.png

    I have seen a few posts which says that as the centripetal velocity is perpendicular to the tangential a change in this force will not result in an increase in the tangential velocity.
    but intuition tells me that as the ball is forced onto a smaller and smaller radius its rotational speed will increase.
    If I thought of this as the components acting upon it - Tangential inertia and Centripetal force then as I increase the force the angle of the resultant vector will decrease .

    upload_2016-9-13_11-35-41.png upload_2016-9-13_11-36-12.png


    Doesn't this mean it has a greater rotational velocity ? doesn't that infer greater tangential velocity ?
     

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  3. Sep 13, 2016 #2

    andrewkirk

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    Your hypothesis is correct. The tangential velocity will increase.

    An easy way to derive this is to note that angular momentum is conserved, because pulling on the string cannot impart any torque on the ball. Angular momentum is ##mr^2\omega=mrv## where ##v## is tangential velocity. Given that this remains constant and ##r## is reduced, ##v## will increase.
     
  4. Sep 13, 2016 #3

    A.T.

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    The force from the string is not perpendicular to velocity when the radius decreases. It has a component parallel to velocity, so it’s doing positive work and increasing the speed of the mass.
     
  5. Sep 13, 2016 #4

    Simon Bridge

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    Where?
    "centripetal velocity" is not a standard term ... a mass in circular motion has components of velocity that are tangential and radial. In uniform circular motion, the radial velocity is zero.
    This argument is puzzling - it seems to be saying that a perpendicular force won't accelerate the object... in this case it just won't produce a torque...

    Initially I thought you were thinking of how you can increase the rotation speed of a stone you twirl on a string by tugging on the string ... but it seems you are thinking in terms or reducing the length of the string by pulling on it. This is the same conservation of momentum effect as an ice skater pulling their arms in to spin faster.

    http://www.webassign.net/question_assets/tamucolphysmechl1/lab_3/manual.html
     
  6. Sep 13, 2016 #5
    Wow Thanks for all your quick replys, much appreciated

    I have found a relevant link that I think means I could even further simplify my example by holding the radius and radial mass distribution constant.

    http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html gives a direct formula relationship -
    cfmag.gif

    Suggesting that increasing the tangential velocity increases the centripetal force

    Would the inverse still hold true ?

    Holding mass and radius constant should increasing the centripetal force increase the tangential velocity ?
     
  7. Sep 13, 2016 #6

    A.T.

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    When you increase the force over time, the radius won't stay constant over time.
     
  8. Sep 13, 2016 #7

    Simon Bridge

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    How would you go about increasing the centripetal force while keeping the radius constant?
    In the case of the mass distribution, where does the centripetal force come from?
     
  9. Sep 13, 2016 #8
    Both very good points, i've struggled with exactly these thoughts , could we avoid a huge longwinded off-topic explanation by asking "Mathematically" ? or am I going to break a rule ? I am aren't I ?
     
  10. Sep 13, 2016 #9

    Simon Bridge

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    You can make a mathematical model do anything... even things that are not physically possible.
    Start from the physics, then do the maths.

    You could also think of it in terms of Newton's laws of motion ... what you are contemplating is that the angular velocity should increase with no change in radius.
    This means there must be an change in the angular momentum over time - ie at some rate. A rate of change of angular momentum is the definition of a torque (like the rate of change of linear momentum is the definition of a force.) Can you apply a torque just by pulling radially on the string?

    You can twirl a rock over your head (horizontally) on the ends of a length of string, changing the speed of the rotation just by how you pull on the string ... but do you only pull radially on the string to do this?
     
  11. Sep 13, 2016 #10

    A.T.

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    You have to decide if you:
    a) compare two scenarios of uniform circular motion (constant radius/speed/force magnitude)
    b) analyse non-circular motion (radius/speed/force magnitude change over time)
     
  12. Sep 13, 2016 #11
    "Tangential velocity" is not a very meaningful construct. Velocity is always tangent to the trajectory. Unless by "tangent" you mean tangent to a different curve and not the trajectory curve. But when we talk about tangential and normal acceleration usually we refer to the trajectory so it would be unusual to have a different reference for velocity. In this sense, there is no normal component of the velocity. You may have a radial component if whatever you call radius is not normal to the trajectory.

    So yes, applying a force you produce an acceleration which means the velocity will change. As the velocity is tangent to trajectory, you can say that the "tangential velocity" changes but this does not bring any extra information. Even for uniform circular motion the "tangential velocity" changes in direction (the tangent changes from point to point).
     
  13. Sep 13, 2016 #12

    andrewkirk

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    One could construct a physical setup where that happened while the radius remained constant, but it would involve tangential acceleration. It can't be done just by pulling on the string.

    If you had a toy aeroplane flying in a circle on the end of a wire at a constant speed, and then using the remote control you increased the power output of the plane's engine, its angular velocity and tangential velocity would increase, and so would the centripetal force in the wire.

    Mind you, a plane being forced to fly in a circle by a string wouldn't fly very well, because it would be constantly yawing. To make it fly smoothly you'd need to bank it, and that would then create a centripetal force of aerodynamic source, thereby invalidating the calculations about tension in the wire.

    But I suggest you stick with your original problem, which is simpler, as all that's needed is to increase the force with which the string is pulled from the centre.
     
  14. Sep 13, 2016 #13
    Thanks for all your responses, Thinking about how to frame my question , I believe I can phrase a scenario in terms of something akin to A and B
    upload_2016-9-14_12-32-27.png
    I have a circular pool table with a red and blue ball, both begin with an equal tangential velocity at point X
    The table and the blue ball are so charged so as to cause a weak local repulsion between them, the red ball holds no charge -
    As such The red ball receives just enough centripetal force to keep it in uniform circular motion whereas the blue ball encounters a greater force deflecting it continually towards the centre. (My attempt to demonstrate a possible greater centripetal force, please don't get hung up on magnetic fields it is just a proxy example)

    As the additional force on the blue ball is constant I could see that the deflection angles will be cumulative from each interaction ( my attempt at resultant force vectors changing ) This causes the ball to travel along a reducing radius r but a larger circumferential difference ω than the red ball . This would agree with our prior example however I have attempted to demonstrate that the Mass distribution hasn't changed by landing the blue ball (after 5 collisions) at the same radius and starting point X to replace the original mass.
    Now if we were replicate the balls from a single example to a large enough N I could see that the average distribution of mass has stayed constant but the vectors of momentum have increased tangentially.

    What am I missing ?
     
  15. Sep 14, 2016 #14

    Simon Bridge

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    The maths ... either you have added up wrong or you do not have a static configuration.
    I cannot tell because the description is not rigorous enough to pin anything down properly (I'd have to make too many guesses).

    I suspect the situation involves some torque, at least temporarily.
    Consider - a solid is modelled as a set of masses connected in a lattice by massless springs.
    Spin it and the springs stretch a little (but not much, because they are very stiff).
    What is happening is that the spin slightly deforms the solid.
    You could change out the springs for stiffer ones ... this increases the available "centripetal force", so the solid does not deform as much from spinning.
    If you tightened the springs wile spinning what happens and why?

    It does not matter what the springs are made of - they could be metal coils, elastic bands, or electrostatic force ...

    Note: it is always possible to produce an arbitrariy complicated setup that you can convince yourself does something non-physical, if you want to get to the physics, you need to simplify. What are you trying to achieve with this?
     
  16. Sep 14, 2016 #15

    A.T.

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    For this to happen your force fields must have applied some net torque.
     
  17. Sep 14, 2016 #16

    jbriggs444

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    As I understand the pool table example, each bounce by the blue ball involves a radial rebound velocity that is greater than its radial impact velocity. It bounces away faster than it arrived. However, this need not also include any change in what we might call tangential velocity. Accordingly, it need not involve any torque or change in angular momentum. Let us proceed on the assumption that this is the intention, that angular momentum is preserved.

    The blue ball will still pass the red ball. But both have the same angular momentum so you might wonder how that could be. The blue ball has a trajectory that carries it closer to the center of the table. Its average angular velocity can be comparatively higher while its angular momentum remains constant. Eventually its trajectory should become a series of close passes by the center, each barely missing to the left. The "centripetal force" required to continue this series of bounces will increase without bound.

    But the limiting trajectory is not circular -- it is anything but. So there is no reason to expect a limiting force of ##\frac{mv^2}{r}##
     
  18. Sep 14, 2016 #17
    Ah I thought we started with the maths ,I was just trying to establish a causal relationship ?
    cfmag.gif

    I think I have found a clue to the torque issue -

    The attached link discusses how a force acting perpendicular changes the resultant vector such that the resultant vector is no longer perpendicular to the original force.

    https://www.physicsforums.com/threa...l-force-change-the-magnitude-of-speed.392768/
    upload_2016-9-15_10-26-29.png

    "In order for a force to change an object's direction of motion, but not its speed, that force must be always perpendicular to the object's velocity. That's not the case in your example. As soon as the wind acts, it changes the plane's velocity so that it is no longer perpendicular to the force."

    If I was to equate that to my red and blue balls this time examining over just an initial interaction and a second resultant interaction

    The red ball being in uniform circular motion only ever receives enough centripetal force to continue this orbit as such it always has a vector perpendicular to the axis of rotation causing the centripetal force to always act towards the center -no tangential component of force no torque

    The blue ball velocity vector Vr, (when its additional displacement is viewed in terms of its secondary interaction) being the resultant vector of the initial is no longer perpendicular to the centripetal force it experiences . As its velocity vector is no longer perpendicular to centripetal force the force is allowed to have a tangential component - torque is provided .
    upload_2016-9-15_10-33-19.png
    How did I do ?
     
  19. Sep 15, 2016 #18

    A.T.

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    The math/physics says nothing about causality here, it's just a quantitative relationship.


    The term "centripetal" is ambiguous when the path becomes non-circular:
    - In the formula above it means: perpendicular to velocity, or towards an instantaneous centre of path curvature at distance r
    - You use it as: towards a fixed centre of some circle, which might have been the initial path
     
  20. Sep 15, 2016 #19
    We have already agreed on this, you will need to read the rest of the thread otherwise we are in danger of going round in circles. For now I really need a 'why' or 'why not' answer.

    I thought that little airplane gives a very good clue so phrase it in terms of vectors and resultants

    No the force is always centripetal towards the centre of rotation - there is absolutely nothing I can do about its vector , the center of mass also does not change , the only vector that changes is the resultant tangential velocity as in the diagram.

    With those things cleared up can you give a clearer answer ? I know that it is very obtuse to ask the inverse of the known 'Velocity increases Centripetal force' but the presence of an additional radial inward acceleration ( not caused by rotation) is what I appear to encounter so I need to now what happens to Velocity if I increase Centripetal force.
     
  21. Sep 15, 2016 #20

    jbriggs444

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    Your most recent diagram and explanation were not understandable. It is not clear what you are claiming or why.
     
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